Root Problem

Algebra Level 2

2 − 3 \sqrt { 2-\sqrt { 3 } } can be expressed as

3 + 1 2 \frac { \sqrt { 3 } +1 }{ \sqrt { 2 } } 3 − 1 2 3- \frac { 1 }{ \sqrt { 2 } } 3 − 1 2 \frac { \sqrt { 3 } -1 }{ \sqrt { 2 } } 2 − 1 3 2-\frac { 1 }{ \sqrt { 3 } }

Manisha Garg by Manisha Garg , 21, India

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1 solution

Manisha Garg
Dec 10, 2015

Multiplying both the numerator and denominator by 2 \sqrt { 2 }

2 − 3 × 2 2 = 2 ( 2 − 3 ) 2 = 4 − 2 3 2 = 3 + 1 − 2 3 2 = ( 3 − 1 ) 2 2 = 3 − 1 2 \quad \frac { \sqrt { 2-\sqrt { 3 } } \times \sqrt { 2 } }{ \sqrt { 2 } } \\ \\ =\quad \frac { \sqrt { 2(2-\sqrt { 3) } } }{ \sqrt { 2 } } \\ =\frac { \sqrt { 4-2\sqrt { 3 } } }{ \sqrt { 2 } } \\ =\frac { \sqrt { 3\quad +\quad 1\quad -2\sqrt { 3 } } }{ \sqrt { 2 } } \\ =\frac { { (\sqrt { \sqrt { 3 } -1 } ) }^{ 2 } }{ \sqrt { 2 } } \\ =\frac { \sqrt { 3 } -1 }{ \sqrt { 2 } }

10 Helpful 0 Interesting 0 Brilliant 0 Confused

Can I ask how do you get (√√3 - 1)^2 / √2 ???

Hon Ming Rou - 5 years, 6 months ago

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using a 2 + b 2 − 2 a b = ( a − b ) 2 { a }^{ 2 }+{ b }^{ 2 }\quad -2ab\quad =\quad { (a }-b)^{ 2 }

we get, 3 2 + 1 2 − 2 3 × 1 = ( 3 − 1 ) 2 { \sqrt { 3 } }^{ 2 }\quad +\quad { \sqrt { 1 } }^{ 2 }\quad -2\sqrt { 3 } \times \sqrt { 1 } \\ \\ =\quad { (\sqrt { 3 } -1) }^{ 2 }

Manisha Garg - 5 years, 6 months ago

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That's very helpful, thanks a lot.

Hon Ming Rou - 5 years, 6 months ago

Did the same way. +1!

Manish Mayank - 5 years, 6 months ago

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